Two variable function on an unbounded set

Question

I have the following function

$$f(x,y) = 3x^2 +5y^2 +2y -4x -2xy + 1$$

I know that $f$ has a critical point in $x=9/14$ and $y=-1/14$. Also, I know that it will be a local minimum. How do I know (or check) that this value is a global minimum, that is, the lowest value of this function?

The domain is unbounded therefore I cannot apply Weierstrass Theorem. Do I count the limits? If yes, which of them? Is there a way to check that this function has a global minimum without even finding a local one?

Answer 1

$f_x = 6x-4 - 2y = 0 = f_y=10y+2-2x \implies 3(10y+2)-4-2y=0\implies28y=-2\implies y = -\dfrac{1}{14}\implies x = \dfrac{9}{14}$ . Thus $\left(\dfrac{9}{14}, -\dfrac{1}{14}\right)$ is the only critical point that is a local minima as confirmed by you. Hence this is also the global minima since the domain is an open unbounded set.

Answer 2

$$ f(x,y) = 3 x^2 + 5 y^2 + 2 y - 4 x - 2 x y - 11 = (p-p_0)^{\top}M(p-p_0)+c $$

with $p = (x,y), p_0 = (9/4,-1/14), c = 173/14$ and $M = \left(\begin{array}{cc} 3 & -1\\ -1 & 5 \end{array}\right)$ with $M > 0$ so $f(x,y)$ has a global minimum.